Bài 8. Tìm x sao cho sqrt (x-2)+sqrt (4-x)=x^2-6x+11

Let the given equation be<br />$$\sqrt{x-2} + \sqrt{4-x} = x^2 - 6x + 11$$<br />The domain requires $2 \le x \le 4$.<br /><br />Let $f(x) = \sqrt{x-2} + \sqrt{4-x}$ and $g(x) = x^2 - 6x + 11$.<br />$f(x)$ is concave, and its maximum occurs at $x=3$, where $f(3) = \sqrt{3-2} + \sqrt{4-3} = 1 + 1 = 2$.<br />$g(3) = 3^2 - 6(3) + 11 = 9 - 18 + 11 = 2$.<br />Since $f(3) = g(3) = 2$, $x=3$ is a solution.<br /><br />Consider $h(x) = f(x) - g(x)$. Then $h(3) = 0$.<br />$h'(x) = \frac{1}{2\sqrt{x-2}} - \frac{1}{2\sqrt{4-x}} - (2x - 6)$.<br />$h'(3) = \frac{1}{2} - \frac{1}{2} - 0 = 0$.<br /><br />Let's examine the second derivatives.<br />$f''(x) = -\frac{1}{4(x-2)^{3/2}} - \frac{1}{4(4-x)^{3/2}} < 0$ for $2 < x < 4$.<br />$g''(x) = 2 > 0$.<br />Thus $h''(x) = f''(x) - g''(x) < 0$ for $2 < x < 4$.<br />Since $h(3) = 0$ and $h''(x) < 0$, $x=3$ is the only solution.<br /><br />Therefore, the solution is $x=3$.<br />